How many bit strings of length 7 either start with the first two bits equal to 11 or end with the last two bits equal to 00 ? Give a specific number (no binomial coefficients or factorials).
To count the number of bit strings of length 7 that start with the first two bits equal to 11, we fix the first two bits and count the number of choices for the remaining five bits. Since the first two bits are fixed, there are $2^5 = 32$ choices for the remaining five bits.
To count the number of bit strings of length 7 that end with the last two bits equal to 00, we fix the last two bits and count the number of choices for the first five bits. Since the last two bits are fixed, there are $2^5 = 32$ choices for the first five bits.
However, we double-counted the bit string 0000000, since it satisfies both conditions. Therefore, we subtract 1 from the total count.
This means that there are $32 + 32 - 1 = \boxed{63}$ bit strings that either start with the first two bits equal to 11 or end with the last two bits equal to 00.